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Question 1 The area under the curve \(y=x^4+3\) from \(x=0\) to \(x=1\) is \(\frac{16}{5}\).
Step 1: Since \(y=x^4+3\) is positive on \([0,1]\), the area under the curve is \(\int_0^1 (x^4+3)\,dx\). Step 2: Split the integral: \(\int_0^1 x^4\,dx+\int_0^1 3\,dx\). Step 3: Find the antiderivative: \(\frac{x^5}{5}+3x\). Step 4: Evaluate from \(0\) to \(1\): \(\left[\frac{x^5}{5}+3x\right]_0^1=\frac{1}{5}+3=\frac{16}{5}\). Step 5: The statement is true. Answer: B. True
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Question 2 The area enclosed between the curves \(y=e^x\) and \(y=e^{-x}\) from \(x=-2\) to \(x=0\) is \(e^2+\frac{1}{e^2}-2\).
Step 1: On \([-2,0]\), \(e^{-x}\) is above \(e^x\). Step 2: The area between the curves is \(\int_{-2}^{0}(e^{-x}-e^x)\,dx\). Step 3: Find the antiderivative: \(\int e^{-x}\,dx=-e^{-x}\), and \(\int e^x\,dx=e^x\). Step 4: Therefore, \(\int(e^{-x}-e^x)\,dx=-e^{-x}-e^x\). Step 5: Evaluate from \(-2\) to \(0\): \([-e^{-x}-e^x]_{-2}^{0}=(-1-1)-(-e^2-e^{-2})\). Step 6: Simplify: \(-2+e^2+e^{-2}=e^2+\frac{1}{e^2}-2\). Step 7: The statement is true. Answer: A. True
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Question 3 The region enclosed by the curves \(y=x\) and \(y=x^2\) is rotated around the x-axis to create a solid. The volume of this solid is \(\frac{3\pi}{15}\) units\(^3\).
Step 1: The curves \(y=x\) and \(y=x^2\) intersect when \(x=x^2\), so \(x=0\) or \(x=1\). Step 2: On \([0,1]\), the upper curve is \(y=x\) and the lower curve is \(y=x^2\). Step 3: Rotating around the x-axis uses the washer method: \(V=\pi\int_0^1\left(R^2-r^2\right)dx\). Step 4: Here \(R=x\) and \(r=x^2\), so \(V=\pi\int_0^1(x^2-x^4)\,dx\). Step 5: Integrate: \(V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1\). Step 6: Substitute \(x=1\): \(V=\pi\left(\frac{1}{3}-\frac{1}{5}\right)=\pi\cdot\frac{2}{15}=\frac{2\pi}{15}\). Step 7: Since \(\frac{2\pi}{15}\neq \frac{3\pi}{15}\), the statement is false. Answer: B. False
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Question 4 The region enclosed by the curves \(y=x\) and \(y=x^2\) is rotated around the line \(x=-2\) to create a solid. The volume of this solid is:
Step 1: The curves \(y=x\) and \(y=x^2\) intersect at \(x=0\) and \(x=1\). Step 2: The region is rotated around the vertical line \(x=-2\), so use the shell method. Step 3: The shell radius is the distance from \(x\) to \(-2\), which is \(x+2\). Step 4: The shell height is upper curve minus lower curve: \(x-x^2\). Step 5: Set up the volume: \(V=\int_0^1 2\pi(x+2)(x-x^2)\,dx\). Step 6: Expand: \((x+2)(x-x^2)=x^2-x^3+2x-2x^2=2x-x^2-x^3\). Step 7: Integrate: \(V=2\pi\int_0^1(2x-x^2-x^3)\,dx=2\pi\left[x^2-\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1\). Step 8: Substitute \(x=1\): \(V=2\pi\left(1-\frac{1}{3}-\frac{1}{4}\right)=2\pi\cdot\frac{5}{12}=\frac{5\pi}{6}\). Answer: B. \(\frac{5\pi}{6}\) units\(^3\)
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